Rationalizations

Before rationalizing it is convenient to simplify the root: $12=4\cdot 3$, therefore $\sqrt{12}=2\sqrt{3}$ and the fraction reduces to $\frac{3}{\sqrt{3}}$.

The denominator is a single root, therefore numerator and denominator are multiplied by $\sqrt{3}$: the product $\sqrt{3}\cdot \sqrt{3}$ equals $3$, hence the denominator becomes rational. The factor $3$ cancels between numerator and denominator and $\sqrt{3}$ remains.

The denominator is a binomial with square roots, therefore numerator and denominator are multiplied by the conjugate binomial $\sqrt{7}+\sqrt{3}$. The product of a difference and the corresponding sum is the difference of the squares, hence the squares of the roots appear in the denominator: $7-3=4$, a rational number.

The factor $4$ in the denominator cancels with the $4$ in the numerator, therefore the result is $\sqrt{7}+\sqrt{3}$.

The radicand is written as a power: $25=5^2$. With a cube root in the denominator, the multiplying factor is the one that completes the cube under the root: since $5^2\cdot 5=5^3$, numerator and denominator are multiplied by $\sqrt[3]{5}$.

The product of cube roots is the cube root of the product, therefore the denominator becomes $\sqrt[3]{5^3}=5$, a rational number. The fraction $\frac{10}{5}$ simplifies to $2$ and the result is $2\sqrt[3]{5}$.